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Asked by purushkumar91422 | 20 Apr, 2021, 09:00: AM

Expert Answer

In first one second period, toy-car's velocity is changed from 0 to 6 m/s .

Hence acceleration a is determined by equation of motion " v = u + a t "

where v is velocity after t second and initial velocity is u

Hence acceleration a = v/t = 6 / 1 = 6 m/s²

Distance S₁travelled in first one second is calculated using equation of motion

" S = ut + (1/2) a t²"

S₁= (1/2) × 6 × 1 × 1 = 3 m

After one second field is reversed , hence acceleration direction is reversed but its magnitude is same

Hence acceleration = -6 m/s²

Since the toy car is moving with velocity 6 m/s at end of 1 second from initial time,

time taken to reduce its velocity to zero is determined as follows

Equation of motion is " v = u - a t "

In above equation of motion v = 0 , hence t = u / a = 6 /6 = 1 s

Distance moved in time duration from t =1 s to t = 2 s is calculated as follows .

S₂ = u t - (1/2) a t²= ( 6 × 1 ) - [ (1/2) × 6 × 1 × 1 ] = 3 m

Since it is given that toy car has moved for 2 seconds after electric field reversed.

We have seen that it has moved in forward direction 1 second after field reversal.

Hence for one second period it will move in backward due to change in direction of acceleration.

Distance S₃ moved in backward direction for 1 second time duration is calculated as follows

S₃= - (1/2) × 6 × 1 × 1 = -3 m

Average velocity = Net displacement / time = ( S₁ + S₂+ S₃) / t = ( 3 + 3 - 3 ) / 3 = 1 m/s

Average speed = Total distance / time = ( S₁+ S₂+ | S₃| ) / t = ( 3+3+3 ) / 3 = 3 m/s

Answered by Thiyagarajan K | 20 Apr, 2021, 10:04: AM

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